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Physics - Electromagnetism

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This is the last question for a final exam preparation sheet and I can't figure out the answer. Any help would be greatly appreciated. Thank you.

- One very long wire carries current 10.0 A to the left along the x axis. A second very long wire carries current 75.0 A to the right along the line
(y = 0.280 m, z = 0).

(a) Where in the plane of the two wires is the total magnetic field equal to zero? (along the y axis)


(b) A particle with a charge of -2.00 µC is moving with a velocity of 150 Mm/s along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle.


(c) What If? A uniform electric field is applied to allow this particle to pass through this region undeflected. Calculate the required vector electric field.

  • Physics - Electromagnetism - ,

    (a)
    Magnetic field of the long current carrying wire according to Biot-Savert Law is
    B=μₒI/2•π•a
    The desired point is separated by distance “x” from the 1st current and by distance (d+x) from the 2nd current (along y-axis).
    μₒI1/2•π•x=μₒI2/2•π•(d+x),
    10•(0.28+x)=75•x
    2.8+10x=75 x.
    65x=2.8
    x=0.431 m.
    (b) Magnetic force (Lorentz force) is
    F=qvB sinα.
    Since the motion of the particle is along the straight line, then F=0
    (c) Electric field has to be applied along the direction of the particle motion (E↑↑ῡ, E↑↓ῡ)

  • Physics - Electromagnetism - ,

    thank you Elena!

  • Physics - Electromagnetism - ,

    For people of the future, Elena's answer for part B is wrong. It's better to use the equation
    F = q*v(cross)B
    where B is the sum of the B field created by the 50 A wire and the 30 A wire.

    ------------------------> 50A
    |
    | .18m
    |
    ------> v = 150E6 m/s
    |
    | .1m
    <-------------------- 30A

    B_tot = mu/2pi * (50/.18 (-k) + 30/.1 (-k))
    = 1.2E-4 (-k) Teslas

    Then you take the cross product and eventually get F = -.035j N

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